Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.7 Solving Quadratic Equations by Factoring - 5.7 Exercise Set: 15

Answer

$x=\left\{ \dfrac{18}{11},\dfrac{1}{21} \right\}$

Work Step by Step

Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ \left( \dfrac{2}{3}x-\dfrac{12}{11} \right)\left( \dfrac{7}{4}x-\dfrac{1}{12} \right) ,$ is \begin{array}{l}\require{cancel} \dfrac{2}{3}x-\dfrac{12}{11}=0 \\\\ \dfrac{2}{3}x=\dfrac{12}{11} \\\\ \dfrac{3}{2}\left(\dfrac{2}{3}x\right)=\left(\dfrac{12}{11}\right)\dfrac{3}{2} \\\\ x=\dfrac{36}{22} \\\\ x=\dfrac{\cancel{2}\cdot18}{\cancel{2}\cdot11} \\\\ x=\dfrac{18}{11} ,\\\\\text{OR}\\\\ \dfrac{7}{4}x-\dfrac{1}{12}=0 \\\\ \dfrac{7}{4}x=\dfrac{1}{12} \\\\ \dfrac{4}{7}\left(\dfrac{7}{4}x\right)=\left(\dfrac{1}{12}\right)\dfrac{4}{7} \\\\ x=\dfrac{4}{84} \\\\ x=\dfrac{\cancel{4}}{\cancel{4}\cdot21} \\\\ x=\dfrac{1}{21} .\end{array} Hence, $ x=\left\{ \dfrac{18}{11},\dfrac{1}{21} \right\} .$
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